Let  and  be the - and -intercepts, respectively, of the line. Then the slope of the line is , or, equilvalently, . 

We do not need to find the actual slopes of the four choices if we observe that in each case,  and  are of the same sign. Since the quotient of two numbers of the same sign is positive, it follows that  is negative, and therefore, none of the pairs of intercepts can be those of a line with positive slope .

 Let  and  be the - and -intercepts, respectively, of the line. Then the slope of the line is , or, equilvalently, . 

We can examine the intercepts in each choice to determine which set meets these conditions.

 and :

Slope: 

 and 

Slope: 

 and 

Slope: 

 and 

Slope: 

 and  comprise the correct choice.

The slope of the green line can be calculated by noting that the - and -intercepts of the line are, respectively,  and .  If  and  be the - and -intercepts, respectively, of a line, the slope of the line is . This makes the slope of the green line . 

Any line perpendicular to this line must have as its slope the opposite reciprocal of this, or . Since the desired line must also have -intercept , the equation of the line, in point=slope form, is

which can be simplified as

To locate the -intercept of the equation , substitute 0 for :

The -intercept of the parabola is .

The vertex of the parabola of an equation of the form  has -coordinate . Here, we substitute , to obtain -coordinate

.

To find the -coordinate, substitute this for :

The vertex is .

The line includes points  and ; apply the slope formula:

The slope is , and the -intercept is ; in the slope-intercept form , substitute for  and . The equation of the line is .

A line with undefined slope is a vertical line, and its equation is  for some , so the -coordinate of all points it passes through is . If it goes through the vertex of a parabola , then the line has the equation . Therefore, all we need to find is the -coordinate of the vertex of the parabola.

The vertex of the parabola of the equation  has as its -coordinate , which, for the parabola of the equation , can be found by setting :

The desired line is .

 Let  and  be the - and -intercepts, respectively, of the line. Then the slope of the line is , or, equilvalently, . 

We can examine the intercepts in each choice to determine which set meets these conditions.

 and 

Slope:  

 and 

Slope:  

 and 

Slope:  

 and 

Slope:  

 and  comprise the correct choice, since a line passing through these points has the correct slope.

The -intercept of the line coincides with that of the graph of the quadratic equation, which is a parabola; to find the -intercept of the parabola, substitute 0 for  in the quadratic equation:

The -intercept of the parabola, and of the line, is .

The -intercept of the line coincides with one of those of the parabola; to find the -intercepts of the parabola, substitute 0 for  in the equation:

Using the  method, split the middle term by finding two integers whose product is  and whose sum is ; by trial and error we find these to be  and 4, so proceed as follows:

Split:

or

The -intercepts of the parabola are  and , so the -intercept of the line is one of these. We examine both possibilities.

If  and  be the - and -intercepts, respectively, of the line, then the slope of the line is , or, equivalently, 

If the intercepts are  and , the slope is ; if the intercepts are  and , the slope is . Since the line is of negative slope, we choose the line of slope ; since its -intercept is , then we can substitute  in the slope-intercept form of the line, , to get the correct equation, .

If  and  be the - and -intercepts, respectively, of a line, the slope of the line is . 

The - and -intercepts of the line are, respectively,  and , so , and consequently, the slope of the green line is .  A line parallel to this line must also have slope . 

Each of the equations of the lines is in slope-intercept form , where  is the slope, so we need only look at the coefficients of . The only choice that has  as its -coefficient is , so this is the correct choice.

The -intercept of the line coincides with that of the graph of the quadratic equation, which is a parabola; to find the -intercept of the parabola, substitute 0 for  in the quadratic equation:

The -intercept of the parabola, and of the line, is .

The -intercept of the line coincides with one of those of the parabola; to find the -intercepts of the parabola, substitute 0 for  in the equation:

The quadratic expression can be "reverse-FOILed" by noting that 9 and  have product  and sum 7:

, in which case  

or

, in which case .

The -intercepts of the parabola are  and , so the -intercept of the line is one of these. We will examine both possibilities

If  and  be the - and -intercepts, respectively, of the line, then the slope of the line is . If the intercepts are  and , the slope is ; if the intercepts are  and , the slope is . Since the line is of positive slope, we choose the line of slope 9; since its -intercept is , then we can substitute  in the slope-intercept form of the line, , to get the correct equation, .

We can find out whether the graphs of  and  intersect by first solving for  in the first equation:

We then substitute in the second equation for :

Then we rewrite in standard form:

Since we are only trying to pdetermine whether at least one point of intersection exists, rather than actually find the point, all we need to do is to evaluate the discriminant; if it is nonnegative, at least one solution - and, consequently, one point of intersection - exists. In the general quadratic equation , this is , so here, the discriminant is

.

Therefore, the line of the equation  intersects the parabola of the equation .

We do the same for the other three lines:

Then we rewrite in standard form:

.

The line of  intersects the parabola.

The line of  intersects the parabola.

Since the discriminant is negative, the system has no real solution. This means that the line of  does not intersect the parabola of the equation , and it is the correct choice.